Question

In an amino acid, the carboxyl group ionizes at $pK_{a_1}$ = 2.34 and ammonium ion at $pK_{a_2}$ = 9.6. The isoelectric point of the amino acid is at pH (Nearest integer)

Answer 1

6

Answer 2

The isoelectric point (pI) of an amino acid is the pH at which the amino acid has no net electrical charge. For a simple amino acid with one carboxyl group and one amino group, the pI is calculated as the average of the pKa values of the carboxyl group and the ammonium ion.

Given: $pK_{a_1}$ (carboxyl group) = 2.34 $pK_{a_2}$ (ammonium ion) = 9.6

The formula for the isoelectric point (pI) is: $pI = \frac{pK_{a_1} + pK_{a_2}}{2}$

Substitute the given values into the formula: $pI = \frac{2.34 + 9.6}{2}$ $pI = \frac{11.94}{2}$ $pI = 5.97$

Rounding the calculated pI to the nearest integer: $5.97 \approx 6$

The isoelectric point of the amino acid is at pH 6.